Question: Let $h(x)=\sqrt{x^5}$. $h'(4)=$
Answer: The strategy We can first rewrite $h(x)$ as a rational power of $x$. Then, the derivative of $h$ can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is a fraction.) Once we have $h'(x)$, we can plug $x=4$ into it to find $h'(4)$. Rewriting the radical as a rational power $h(x)=\sqrt{x^5}=x^{^{\frac{5}{2}}}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}h'(x) \\\\ &=\dfrac{d}{dx}\left(x^{^{\frac{5}{2}}}\right) \\\\ &=\dfrac{5}{2}x^{^{\frac{5}{2}-1}} \gray{\text{The power rule}} \\\\ &=\dfrac52x^{^{\frac{3}{2}}} \end{aligned}$ Evaluating $h'(x)$ So we found that $h'(x)=\dfrac52x^{^{\frac{3}{2}}}$, which can also be written as $2.5{\sqrt{ x^3}}$. Now let's plug ${x=4}$ : $\begin{aligned} 2.5\sqrt{({4})^3}&=2.5\sqrt{64} \\\\ &=2.5\cdot 8 \\\\ &=20 \end{aligned}$ In conclusion, $h'(4)=20$.